Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class Q>
Q _sqr(Q x) {
  return x * x;
}
template <class Q>
void gi(Q &x) {
  char ch = getchar();
  x = 0;
  int s = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') s = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - 48, ch = getchar();
  if (s == -1) x = -x;
}
template <class Q>
void pi(Q x) {
  if (x < 0) {
    putchar('-');
    pi(-x);
  } else {
    if (x > 9) pi(x / 10);
    putchar(x % 10 + 48);
  }
}
const int sz = 400200;
const int huge = 1000000000;
const long long int hugell = 1ll << 61;
const double hugelf = 1e50;
const double eps = 1e-10;
const int ljz = 1000000007;
void MOD(int &x) {
  if (x >= ljz) x -= ljz;
  if (x < 0) x += ljz;
}
int Tm(int a, int b) {
  long long int x = a, y = b;
  return (x * y) % ljz;
}
int fpm(int x, int y) {
  int p = 1, s = x;
  for (; y; y >>= 1) {
    if (y & 1) p = Tm(p, s);
    s = Tm(s, s);
  }
  return p;
}
int Tm(int a, int b, int c) { return Tm(Tm(a, b), c); }
int Tm(int a, int b, int c, int d) { return Tm(Tm(a, b, c), d); }
int Tm(int a, int b, int c, int d, int e) { return Tm(Tm(a, b, c, d), e); }
int Tm(int a, int b, int c, int d, int e, int f) {
  return Tm(Tm(a, b, c, d, e), f);
}
int n, m, a[233][233], c[233], d[233], i, j, k;
int main() {
  scanf("%d %d", &n, &m);
  for (i = 1; i <= n; i++)
    for (j = 1; j <= m; j++) scanf("%d", &a[i][j]);
  for (i = 1; i <= n; i++)
    for (j = 1; j <= m; j++)
      for (k = 1; k <= m; k++)
        if (a[i][j] - a[i][k] != a[1][j] - a[1][k]) return printf("-1"), 0;
  if (n < m) {
    for (i = 1; i <= n; i++) {
      c[i] = 1000000;
      for (j = 1; j <= m; j++) c[i] = min(c[i], a[i][j]);
    }
    for (i = 1; i <= m; i++) d[i] = a[1][i] - c[1];
  } else {
    for (i = 1; i <= m; i++) {
      d[i] = 1000000;
      for (j = 1; j <= n; j++) d[i] = min(d[i], a[j][i]);
    }
    for (i = 1; i <= n; i++) c[i] = a[i][1] - d[1];
  }
  for (i = 1; i <= n; i++)
    for (j = 1; j <= m; j++)
      if (c[i] + d[j] != a[i][j]) return printf("-1"), 0;
  k = 0;
  for (i = 1; i <= n; i++) k += c[i];
  for (i = 1; i <= m; i++) k += d[i];
  printf("%d\n", k);
  for (i = 1; i <= n; i++)
    while (c[i]--) printf("row %d\n", i);
  for (i = 1; i <= m; i++)
    while (d[i]--) printf("col %d\n", i);
  return 0;
}